To find the value of $$n$$ in the expression $$1000! = 3^n \times m$$ where $$m$$ is an integer not divisible by 3, we need to determine the largest power of 3 that divides $$1000!$$. This can be accomplished by using Legendre's formula, which is also known as de Polignac's formula. The formula is used to determine the exponent of a prime number $$p$$ in the factorization of $$n!$$.

Legendre's formula is given by:

$$ e_p(n!) = \left\lfloor \frac{n}{p} \right\rfloor + \left\lfloor \frac{n}{p^2} \right\rfloor + \left\lfloor \frac{n}{p^3} \right\rfloor + \left\lfloor \frac{n}{p^4} \right\rfloor + \cdots $$

where $$\left\lfloor x \right\rfloor$$ represents the floor function, which gives the greatest integer less than or equal to $$x$$, and $$p$$ is the prime number we are interested in (in this case, 3).

We need to find:

$$ e_3(1000!) = \left\lfloor \frac{1000}{3} \right\rfloor + \left\lfloor \frac{1000}{3^2} \right\rfloor + \left\lfloor \frac{1000}{3^3} \right\rfloor + \left\lfloor \frac{1000}{3^4} \right\rfloor + \cdots $$

Let's calculate each term step by step:

$$ \left\lfloor \frac{1000}{3} \right\rfloor = \left\lfloor 333.3333 \right\rfloor = 333 $$

$$ \left\lfloor \frac{1000}{9} \right\rfloor = \left\lfloor 111.1111 \right\rfloor = 111 $$

$$ \left\lfloor \frac{1000}{27} \right\rfloor = \left\lfloor 37.037 \right\rfloor = 37 $$

$$ \left\lfloor \frac{1000}{81} \right\rfloor = \left\lfloor 12.3457 \right\rfloor = 12 $$

$$ \left\lfloor \frac{1000}{243} \right\rfloor = \left\lfloor 4.115 \right\rfloor = 4 $$

$$ \left\lfloor \frac{1000}{729} \right\rfloor = \left\lfloor 1.371 \right\rfloor = 1 $$

For powers higher than 3^6, the terms will be zero since 1000 divided by any higher powers of 3 will be less than 1.

Summing these values, we get:

$$ e_3(1000!) = 333 + 111 + 37 + 12 + 4 + 1 = 498 $$

Therefore, the value of $$n$$ is:

$$ n = 498 $$

The correct option is:

Option A

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